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Math Help - I got some crazy simultaneous equation that no one can solve

  1. #1
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    I got some crazy simultaneous equation that no one can solve

    Here it is:

    (a + b)x + cy = bc

    (b + c)y + ax = -ab

    Find x and y.

    I've tried this for ages but the answers are  x= c and y=-a

    Please help
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  2. #2
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    In matrix form this would be

    \displaystyle \left[\begin{matrix}a+b&c\\b+c&a\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}bc\\8722ab\end{matrix}\right]

    \displaystyle \left[\begin{matrix}x\\y\end{matrix}\right] = \left[\begin{matrix}a+b&c\\b+c&a\end{matrix}\right]^{-1}\left[\begin{matrix}bc\\8722ab\end{matrix}\right].


    Can you go from here?
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  3. #3
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    By the substitution method:

    [1] \,\,\, (a + b)x + cy = bc

    [2] \,\,\, (b + c)y + ax = -ab

    [2] \,\,\, y = \dfrac{-ab - ax}{b + c}

    Substitute into [1]:

    [1] \,\,\, (a + b)x + \dfrac{c(-ab - ax)}{b + c} = bc

    And the simplify to solve for x. Place the value of x back into one of the equations to solve for y.
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  4. #4
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    Hello, Prove it!

    Be careful . . . The original equations were "scrambled" . . .


    \begin{array}{ccc}(a + b)x + cy &=& bc \\<br />
ax + (b + c)y &=& \text{-}ab\end{array}\qquad \text{(unscrambled form)}

    We want: . \begin{bmatrix}a+b & c \\ a & b+c\end{bmatrix}\,\begin{bmatrix} x \\ y\end{bmatrix} \;=\;\begin{bmatrix}bc \\ \text{-}ab\end{bmatrix}

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  5. #5
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    WOW!!! I GOT IT USING MATRICES!!!!!!!

    WOOOOOOOOOOOOOOOOOOOOOOOO

    Thanks Prove It and Soroban

    I have too much working to type it all but the determinant is (a+b)(b+c)-ac Then just simplify away.

    thanks!!!!
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  6. #6
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    Quote Originally Posted by jgv115 View Post
    (a + b)x + cy = bc
    (b + c)y + ax = -ab
    I've tried this for ages but the answers are  x= c and y=-a
    Huh? Try a = 2, b = 4, c = -6, x = 1, y = 5

    That's a solution for both equations.....but x<>c and y<>-a
    There are many other similar cases.

    Did I miss something?

    Also, equations can be re-arranged this way:
    ax + bx + cy = bc [1]
    -ax - by - cy = ab [2]
    ===============
    bx - by = ab + bc
    x - y = a + c
    Last edited by Wilmer; November 20th 2010 at 04:51 AM.
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